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A man is standing on top of a building 100 m high. He throws two ball vertically, one at `t=0` and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At `t=2`, both the balls reach to their maximum heights at this time the vertical gap between first and second ball is `+15m`.
Q. The speed of first ball is

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To solve the problem, we will follow these steps: ### Step 1: Define the variables Let the speed of the first ball be \( v_1 \) and the speed of the second ball be \( v_2 \). According to the problem, \( v_2 = \frac{1}{2} v_1 \). ### Step 2: Determine the maximum heights The maximum height reached by an object thrown vertically upwards can be calculated using the formula: \[ ...
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